MHB Convergence of $\displaystyle\sum\frac{n^5}{2^n}$

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The series $\sum\frac{n^5}{2^n}$ converges, as determined by applying the ratio test and the root test. The limit of the ratio test yields $\frac{1}{2}$, indicating convergence. Additionally, the root test confirms that the series converges for all values of $k$. Thus, both tests validate the convergence of the series. The discussion emphasizes the effectiveness of these tests in determining convergence.
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Does the following series converge?

$\displaystyle\sum\frac{n^5}{2^n}$
 
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Well what tests did you try?the ratio test and the root test both give you what you need. Try them. If you need more hints let us know.Mohammad
 
fawaz said:
Well what tests did you try?the ratio test and the root test both give you what you need. Try them. If you need more hints let us know.Mohammad

$\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{(n+1)^5}{2n^5}=\frac{1}{2}$

So, $\displaystyle\sum\frac{n^5}{2^n}$ converges.
 
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$a_n=\dfrac{n^k}{2^n},$ so by the root test the series always converges for all $k.$
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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